Question

The equation of three circles are given `x^2+y^2=1,x^2+y^2-8x+15=0,x^2+y^2+10 y+24=0` . Determine the coordinates of the point `P` such that the tangents drawn from it to the circle are equal in length.

Answer 1

We known that the point from which the lengths of tangents are equal in length is the radical of the given three circles. Now, the radical axis of the first two circles is
`(x^(2)+y^(2)-1)-(x^(2)+y^(2)-8x+15)=0`
`i.e., x-2=0` (1)
and the radical axis of the second and third circle is
`(x^(2)+y^(2)-8x+15)-(x^(2)+y^(2)+10y+24)=0`
i.e., `8x+10y+9=0` (2)
Solving (1) and (2), the coordinates of the radical center, i.e., of the point P, are `P(2,-5//2)`.